Integral Valuation of Continuous Whole-Life Insurance

Why the Integral Valuation Matters

A continuous whole-life policy pays its benefit at the moment of death, rather than at the end of the policy year of death. The time of payment can be any nonnegative real number, so the present value must account for both how likely death is at each time and how much a payment at that time is discounted. The integral valuation formula brings these two ingredients together.

The central task is to turn a survival model into a probability density for the time of death, weight that density by the discount factor at each possible death time, and integrate over all possible times. This approach works for many survival models; it does not require mortality to have a constant force.

Definition: Let $T_x$ be the future lifetime of a life aged $x$, and let ${}_tp_x=P(T_x>t)$ be the probability of surviving at least $t$ more years. A unit continuous whole-life insurance pays 1 immediately at death. Under a constant force of interest $\delta$, its present value at issue is $Z=e^{-\delta T_x}$. Its expected present value is denoted $\bar A_x$.

From Survival to a Death-Time Density

Assume future lifetime has an absolutely continuous distribution, with survival probability ${}_tp_x$ differentiable in $t$. The force of mortality at future duration $t$ is $\mu_{x+t}$. It describes the instantaneous rate of death at duration $t$, conditional on surviving to that duration. In particular, over a short interval of length $h$, the conditional probability of death is approximately $\mu_{x+t}h$.

Definition: Under the stated continuous-lifetime assumption, the density of $T_x$ at duration $t$ is $$ f_{T_x}(t)={}_tp_x\mu_{x+t}=-\frac{d}{dt}{}_tp_x. $$ The survival probability accounts for reaching duration $t$; the force of mortality accounts for dying at that duration, conditional on reaching it.

This identity also follows by differentiating the survival function. Since the force of mortality is $\mu_{x+t}=-\frac{d}{dt}\ln({}_tp_x)$, the derivative of survival is $\frac{d}{dt}{}_tp_x=-{}_tp_x\mu_{x+t}$. Thus its negative derivative, which is the density, is ${}_tp_x\mu_{x+t}$.

For constant force of interest $\delta$, the discount factor for a payment $t$ years from now is $e^{-\delta t}$. This convention discounts continuously over fractional years. If an effective annual interest rate $i$ is instead specified and fractional-year discounting is based on an equivalent constant force, use $\delta=\ln(1+i)$. The mortality force $\mu_{x+t}$ and the interest force $\delta$ are distinct quantities.

The Whole-Life Integral Formula

Theorem: If death occurs with probability 1, future lifetime has an absolutely continuous distribution, and the force of interest is constant at $\delta$, then the expected present value of a unit continuous whole-life insurance is $$ \bar A_x =E[e^{-\delta T_x}] =\int_0^\infty e^{-\delta t}f_{T_x}(t)\,dt =\int_0^\infty e^{-\delta t}{}_tp_x\mu_{x+t}\,dt. $$

If the force of interest varies with time, let $\delta(s)$ be the force at time $s$, and define the discount function $D(t)=\exp\left(-\int_0^t\delta(s)\,ds\right)$. The same argument gives $\bar A_x=\int_0^\infty D(t){}_tp_x\mu_{x+t}\,dt$, subject to the same continuous-lifetime assumptions. The constant-force formula is the special case $D(t)=e^{-\delta t}$.

Write the present value at a possible death time.
If death occurs at duration $t$, the unit benefit has present value $e^{-\delta t}$.

Weight by the probability density of that death time.
The density is ${}_tp_x\mu_{x+t}$, not the force of mortality alone.

Accumulate contributions over all possible death times.
Integrating the product from 0 to infinity gives the expected present value.

The derivation is the continuous version of an expected-value calculation: for a random variable with density $f$, the expectation of a function $g(T_x)$ is $\int_0^\infty g(t)f(t)\,dt$. Set $g(t)=e^{-\delta t}$ and $f(t)={}_tp_x\mu_{x+t}$. The formula follows directly. For a benefit amount $B$ paid at death, the present value and expected present value are multiplied by $B$.

The density formula assumes there is no probability mass at an isolated time of death. If the model allows an atom—for example, a positive probability of death at a specified instant—the integral over the continuous density alone omits that probability. Its discounted contribution must be added separately. Likewise, if some lives never die under the model, the density need not integrate to 1; the formula still values payments on deaths, but does not imply that every policy produces a payment.

Worked Example: An Erlang Future-Lifetime Model

A hypothetical insurer models a life aged $x$ with survival function ${}_tp_x=e^{-t}(1+t)$ for $t\geq 0$. Future lifetime follows an Erlang distribution with shape 2 and rate 1 per year. The force of interest is $\delta=1$ per year. Find $\bar A_x$. Use the integral valuation method and identify the common mistake it avoids.

Step 1: Find the force of mortality. Differentiate the survival function: $$ \frac{d}{dt}\left[e^{-t}(1+t)\right] =-te^{-t}. $$ Therefore the density is $f_{T_x}(t)=te^{-t}$. Dividing by survival gives $$ \mu_{x+t}=\frac{te^{-t}}{e^{-t}(1+t)}=\frac{t}{1+t}. $$ The survival probability times the force is indeed $e^{-t}(1+t)\frac{t}{1+t}=te^{-t}$.

Step 2: Apply the interest discount. With $\delta=1$, the discount factor at duration $t$ is $e^{-t}$. Hence $$ \bar A_x=\int_0^\infty e^{-t}te^{-t}\,dt =\int_0^\infty te^{-2t}\,dt. $$

Step 3: Evaluate the integral. Integration by parts, with $u=t$ and $dv=e^{-2t}dt$, gives $$ \int_0^\infty te^{-2t}\,dt =\left[-\frac{t}{2}e^{-2t}\right]_0^\infty +\frac12\int_0^\infty e^{-2t}\,dt =0+\frac12\left(\frac12\right)=\frac14. $$ Thus $\bar A_x=0.2500$, to four decimal places.

Check and mistake avoided. The density integrates to 1 because $\int_0^\infty te^{-t}dt=1$, consistent with eventual death. The method avoids treating $\mu_{x+t}$ alone as a density: the force is conditional on survival, so omitting ${}_tp_x$ would give the wrong expected value.

Worked Example: A Finite-Support Survival Model

For a hypothetical insured life, suppose ${}_tp_x=(1-t/2)^2$ for $0\leq t\leq2$, and ${}_tp_x=0$ for $t>2$. The force of interest is $\delta=1$ per year. Find the expected present value of a unit benefit paid at death.

Step 1: Obtain the death-time density. For $0\leq t<2$, $$ f_{T_x}(t)=-\frac{d}{dt}(1-t/2)^2=1-t/2. $$ Equivalently, $\mu_{x+t}=(1-t/2)^{-1}$ for $t<2$, so survival times force gives $1-t/2$. The force becomes unbounded as $t$ approaches 2, but the density remains integrable. The survival function reaches zero at 2, so there is no probability mass remaining beyond that time.

Step 2: Set the correct integration range. Death occurs by duration 2 with probability 1, and the density is zero outside that range. Therefore $$ \bar A_x=\int_0^2 e^{-t}(1-t/2)\,dt =\int_0^2 e^{-t}\,dt-\frac12\int_0^2 te^{-t}\,dt. $$

Step 3: Evaluate both terms. $$ \int_0^2 e^{-t}\,dt=1-e^{-2}, \qquad \int_0^2 te^{-t}\,dt =\left[-(t+1)e^{-t}\right]_0^2 =1-3e^{-2}. $$ Thus $$ \bar A_x=(1-e^{-2})-\frac12(1-3e^{-2}) =\frac12+\frac12e^{-2} \approx 0.5677. $$ The answer is $0.5677$, to four decimal places.

Check and mistake avoided. The density integrates to $\int_0^2(1-t/2)dt=1$, as required. Integrating to infinity while continuing the polynomial density beyond 2 would incorrectly assign probability to impossible death times; the survival model determines the integration range.

Worked Example: A Mixture of Lifetime Distributions

A hypothetical mortality model assigns probability 0.40 to an exponential lifetime with rate $0.5$ per year and probability 0.60 to an exponential lifetime with rate $1.5$ per year. The mixture is selected at issue and is not observed by the insurer. The force of interest is $\delta=0.5$ per year. Find the expected present value of a \$80,000 whole-life benefit paid at death.

Step 1: Form the mixture survival function. Conditional on the selected component, the survival functions are $e^{-0.5t}$ and $e^{-1.5t}$. Averaging with the stated probabilities gives $$ {}_tp_x=0.40e^{-0.5t}+0.60e^{-1.5t}. $$

Step 2: Differentiate to find the unconditional density. $$ f_{T_x}(t)=0.20e^{-0.5t}+0.90e^{-1.5t}. $$ This density integrates to $0.20/0.5+0.90/1.5=0.40+0.60=1$, confirming that it represents a proper lifetime distribution. The mixture's force is the ratio $f_{T_x}(t)/{}_tp_x$; it is not either component's force at every duration.

Step 3: Discount and integrate. $$ \bar A_x =\int_0^\infty e^{-0.5t} \left(0.20e^{-0.5t}+0.90e^{-1.5t}\right)dt =\frac{0.20}{1.0}+\frac{0.90}{2.0} =0.6500. $$ The expected present value of the \$80,000 benefit is $80{,}000(0.6500)=\$52{,}000.00$.

Check and mistake avoided. The factor 0.6500 lies between 0 and 1, as a discounted unit benefit should when $\delta\geq0$. Applying one component's mortality force to the entire population would ignore the mixture weights and generally produce the wrong value.

Interpretation and Common Exam Traps

The integrand $e^{-\delta t}{}_tp_x\mu_{x+t}$ can be read as the contribution to expected present value from deaths around duration $t$. A higher probability of surviving to $t$ increases the number of lives exposed to death at that duration; a higher force increases the conditional chance of death there; discounting reduces the value of a payment made later. The integral adds these contributions across the lifetime.

A useful check is to consider nonnegative interest. Since $0<e^{-\delta t}\leq1$ when $\delta\geq0$, the expected present value of a unit benefit cannot exceed the probability that a benefit is paid. If death occurs with probability 1, then $0<\bar A_x\leq1$. A result above 1 usually signals an error in the density, the discount factor, or the integration.

A common incorrect calculation replaces death at the exact time $T_x$ by payment at the end of the year of death. For example, using $v^k$ for a death in year $k+1$ values an end-of-year payment, not a continuous benefit. The continuous formula uses $e^{-\delta t}$ at each exact duration $t$. Another error is to integrate $e^{-\delta t}\mu_{x+t}$ without survival. Since $\mu_{x+t}$ is conditional on being alive at duration $t$, it must be multiplied by ${}_tp_x$ to obtain the unconditional death-time density.

The integral method is especially useful when a survival function or force of mortality is specified directly, and when the death benefit is paid continuously at death. Its essential order is consistent: identify the discount function, obtain the death-time density from survival and mortality, and integrate over the support of the lifetime distribution. This tutorial is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the continuous whole-life integral and the stated force of interest.

A hypothetical lifetime has survival function ${}_tp_x=0.25e^{-t}+0.75e^{-2t}$ for $t\geq0$. If $\delta=1$ per year, find the expected present value of a unit benefit paid at death. (Answer: $0.6250$)
A hypothetical lifetime is uniform on $[0,2]$, so ${}_tp_x=1-t/2$ for $0\leq t\leq2$ and zero thereafter. With $\delta=\ln 2$ per year, find the expected present value of a unit benefit. (Answer: $3/(8\ln 2)\approx0.5410$)
A hypothetical lifetime has survival function ${}_tp_x=e^{-t}(1+t)$. The force of interest is $\delta=0.5$ per year. Find the expected present value of a unit benefit, showing the death-time density used. (Answer: $4/9\approx0.4444$)
A hypothetical mixture has survival function ${}_tp_x=0.30e^{-t}+0.70e^{-2t}$, with mixture weights 0.30 and 0.70. The force of interest is $\delta=0.5$ per year. Find the expected present value of a \$25,000 benefit paid at death. (Answer: $\$19{,}000.00$)
A hypothetical lifetime has survival function ${}_tp_x=0.50e^{-t}+0.50e^{-3t}$, and the force of interest is $\delta=1$ per year. Find both the expected present value of a unit benefit and the force of mortality at $t=0.5$. (Answer: $\bar A_x=0.6250$; $\mu_{x+0.5}=(e+3)/(e+1)\approx1.537$ per year)

What You'll Learn